1. Solving equations involving exponents and logarithms

2. Let’s review some terms
Let’s review some terms. When we write log is called the base 125 is called the argument

3. Logarithmic form of 52 = 25 is log525 = 2

4. For all the laws a, M and N > 0
r is any real

5. ln is a short cut for loge log means log10
Remember ln and log
ln is a short cut for loge
log means log10

6. Log laws

7. If your variable is in an exponent or in the argument of a logarithm
Find the pattern your equation resembles

8. If your variable is in an exponent or in the argument of a logarithm
Find the pattern
Fit your equation to match the pattern
Switch to the equivalent form
Solve the result
Check (be sure you remember the domain of a log)

9. log(2x) = 3

10. log(2x) = 3
It fits

11. log(2x) = 3
103=2x
Switch
Did you remember that log(2x) means log10(2x)?

12. log(2x) = 3
103=2x
500 = x
Divide by 2

13. ln(x+3) = ln(-7x)

14. ln(x+3) = ln(-7x)
It fits

15. ln(x+3) = ln(-7x)
Switch

16. ln(x+3) = ln(-7x)
x + 3 = -7x
Switch

17. ln(x+3) = ln(-7x)
x + 3 = -7x
x = - ⅜
Solve the result
(and check)

18. ln(x) + ln(3) = ln(12)

19. ln(x) + ln(3) = ln(12)
x + 3 = 12

20. ln(x) + ln(3) = ln(12)
x + 3 = 12
Oh NO!!! That’s wrong!

21. ln(x) + ln(3) = ln(12)
ln(3x) = ln (12)
You need to use log laws

22. ln(x) + ln(3) = ln(12)
ln(3x) = ln (12)
3x = 12
Switch

23. ln(x) + ln(3) = ln(12)
ln(3x) = ln (12)
3x = 12
x = 4
Solve the result

24. log3(x+2) + 4 = 9

25. log3(x+2) + 4 = 9
It will fit

26. log3(x+2) + 4 = 9
log3(x+2) = 5
Subtract 4 to make it fit

27. log3(x+2) + 4 = 9
log3(x+2) = 5
Switch

28. log3(x+2) + 4 = 9
log3(x+2) = 5
35 = x + 2
Switch

29. log3(x+2) + 4 = 9
log3(x+2) = 5
35 = x + 2
x = 241
Solve the result

30. 5(10x) = 19.45

31. 5(10x) = 19.45
10x = 3.91
Divide by 5 to fit

32. 5(10x) = 19.45
10x = 3.91
Switch

33. 5(10x) = 19.45
10x = 3.91
log(3.91) = x
Switch

34. 5(10x) = 19.45
10x = 3.91
log(3.91) = x
≈ 0.592
Exact log(3.91) Approx 0.592

35. 2 log3(x) = 8

36. 2 log3(x) = 8
It will fit

37. 2 log3(x) = 8
log3(x) = 4
Divide by 2 to fit

38. 2 log3(x) = 8
log3(x) = 4
Switch

39. 2 log3(x) = 8
log3(x) = 4
34=x
Switch

40. 2 log3(x) = 8
log3(x) = 4
34=x
x = 81
Then Simplify

41. log2(x-1) + log2(x-1) = 3

42. log2(x-1) + log2(x-1) = 3
Need to use a log law

43. log2(x-1) + log2(x+1) = 3
log2{(x-1)(x+1)} = 3

44. log2(x-1) + log2(x+1) = 3
log2{(x-1)(x+1)} = 3
Switch

45. log2(x-1) + log2(x+1) = 3
log2{(x-1)(x+1)} = 3
23=(x-1)(x+1)
Switch

46. log2(x-1) + log2(x+1) = 3
log2{(x-1)(x+1)} = 3
23=(x-1)(x+1) = x2 -1
x = +3 or -3
and finish

47. log2(x-1) + log2(x+1) = 3
log2{(x-1)(x+1)} = 3
23=(x-1)(x+1) = x2 -1
x = +3 or -3
But -3 does not check!

48. Exclude -3 (it would cause you to have a negative argument)
log2(x-1) + log2(x+1) = 3
log2{(x-1)(x+1)} = 3
23=(x-1)(x+1) = x2 -1
x = +3 or -3
Exclude (it would cause you to have a negative argument)

49. There’s more than one way to do this

50. Can you find why each step is valid?

51. rules of exponents multiply both sides by 2
- 3 to get exact answer
Approximate answer

52. Here’s another way to solve the same equation.

53. Square both sides
Simplify
exclude 2nd result

54. 52x - 5x – 12 = 0

55. Factor it. Think of y2 - y-12=0
52x - 5x – 12 = 0
(5x – 4)(5x + 3) = 0
Factor it. Think of y2 - y-12=0

56. 52x - 5x – 12 = 0
(5x – 4)(5x + 3) = 0
5x – 4 = 0 or 5x + 3 = 0
Set each factor = 0

57. Solve first factor’s equation
Solve 5x – 4 = 0
5x = 4
log54 = x
Solve first factor’s equation

58. Solve other factor’s equation
Solve 5x + 3 = 0
5x = -3
log5(-3) = x
Solve other factor’s equation

59. Oops, we cannot have a negative argument
Solve 5x + 3 = 0
5x = -3
log5(-3) = x
Oops, we cannot have a negative argument

60. Only the other factor’s solution works
Solve 5x + 3 = 0
5x = -3
log5(-3) = x
Exclude this solution.
Only the other factor’s solution works

61. 4x+2 = 5x

62. 4x+2 = 5x
If M = N then ln M = ln N

63. 4x+2 = 5x
ln(4x+2) = ln(5x )
If M = N then ln M = ln N

64. 4x+2 = 5x
ln(4x+2) = ln(5x)

65. 4x+2 = 5x
ln(4x+2) = ln(5x)
(x+2)ln(4) =(x) ln(5 )

66. Distribute 4x+2 = 5x ln(4x+2) = ln(5x) (x+2)ln(4) =(x) ln(5 )
x ln (4) + 2 ln(4) = x ln (5)
Distribute

67. Get x terms on one side 4x+2 = 5x ln(4x+2) = ln(5x)
x ln (4) + 2 ln(4) = x ln (5)
x ln (4) - x ln (5)= 2 ln(4)
Get x terms on one side

68. Factor out x 4x+2 = 5x ln(4x+2) = ln(5x) (x+2)ln(4) =(x) ln(5 )
x ln (4) + 2 ln(4) = x ln (5)
x ln (4) - x ln (5)= 2 ln(4)
x [ln (4) - ln (5)]= 2 ln(4)
Factor out x

69. Divide by numerical coefficient
4x+2 = 5x
ln(4x+2) = ln(5x)
(x+2)ln(4) =(x) ln(5 )
x ln (4) + 2 ln(4) = x ln (5)
x ln (4) - x ln (5)= 2 ln(4)
x [ln (4) - ln (5)]= 2 ln(4)
Divide by numerical coefficient